Which integral gives the arc length of the curve $y=\ln(x)$ over the interval $[1, \sqrt3]$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $ \int_{1}^{\sqrt3} \sqrt{1+\ln x}\,dx$ (Choice B) B $ \int_{1}^{\sqrt3} \sqrt{1+\ln^2x}\,dx$ (Choice C) C $ \int_{1}^{\sqrt3} \sqrt{1+\dfrac1{x^2}}\,dx$ (Choice D) D $ \int_{1}^{\sqrt3} \sqrt{1+\dfrac1x}\,dx$
Explanation: The arc length $L$ of the curve $y=f(x)$ over the interval $[a, b]$ is $ L=\int_a^b\sqrt{1+\left(\dfrac{dy}{dx}\right)^2}\, dx$. First, calculate $dy/dx$. $\begin{aligned} y &= \ln x\\ \\ \dfrac{dy}{dx} &= \dfrac1x \end{aligned}$ Now apply the arc length formula on the interval $[1, \sqrt3]$ and simplify the integral. $\begin{aligned} L &= \int_{1}^{\sqrt3} \sqrt{1+\left(\dfrac1{x}\right)^2}\,dx \\\\ &= \int_{1}^{\sqrt3} \sqrt{1+\dfrac1{x^2}}\,dx \end{aligned}$